Quantum - Electron in an infinite rectangular prism well

[golmschenk]

Homework Statement

If an electron is in an infinite rectangular prism well, with sides of length a, b, and c where c is the shortest and (b^2)*c=a^3, for what value of the d=b/c is the first excited state of the electron minimized? This isn't the complete problem but it's the part that's giving me trouble/

Homework Equations

I'm using the equations E=(h^2*pi^2/(2m))*((n_x/l_x)^2*(n_y/l_y)^2*(n_z/l_z)^2) but it's suppose to be used for an electron gas in a solid. The question is referring to a crystal. Is this equation one I want to use? Sorry for not using the correct math notation to make it look nice.

The Attempt at a Solution

My attempted solution is basically just plugging it into that equation.

Thanks for your time.

 

[lanedance]

Homework Statement

If an electron is in an infinite rectangular prism well, with sides of length a, b, and c where c is the shortest and (b^2)*c=a^3, for what value of the d=b/c is the first excited state of the electron minimized? This isn't the complete problem but it's the part that's giving me trouble/

Homework Equations

I'm using the equations E=(h^2*pi^2/(2m))*((n_x/l_x)^2*(n_y/l_y)^2*(n_z/l_z)^2) but it's suppose to be used for an electron gas in a solid. The question is referring to a crystal. Is this equation one I want to use? Sorry for not using the correct math notation to make it look nice.

Thanks for your time.

i'm not that familiar with the equation you give, but don't think its the the correct one to use

the infinite potential rectangular box has a reasonable analytic solution solved through separation of variables (the spatial cartesian variables can be separated in the DE )

you will get a similar equation to the one you quote for energy, however the \frac({n_x}{L_x})^2 terms are summed not multiplied

 

[golmschenk]

Oops, yeah, summed is actually what I meant to type. And I think that ended up working out. I don't know for sure that I got the right answer yet, but it seemed to work. Thanks.